Factor the following expression: $ x^2 + xy - 12y^2 $
When we factor a polynomial of this form, we are basically reversing this process of multiplying linear expressions together: $ \begin{eqnarray} (x + ay)(x + by)&=&xx &+& xby + ayx &+& ayby \\ \\ &=& x^2 &+& {(a+b)}xy &+& {ab}y^2 \\ &\hphantom{=}& \hphantom{x^2} &\hphantom{+}& \hphantom{{1}xy} &\hphantom{+}& \hphantom{{-12}y^2} \end{eqnarray} $ $ \begin{eqnarray} \hphantom{(x + ay)(x + by)}&\hphantom{=}&\hphantom{xx} &\hphantom{+}& \hphantom{xby + ayx} &\hphantom{+}&\hphantom{ayby} \\ &\hphantom{=}& \hphantom{x^2} &\hphantom{+}&\hphantom{{(a+b)}xy}&\hphantom{+}&\hphantom{{ab}y^2} \\ &=& x^2 &+& {1}xy &+& {-12}y^2 \end{eqnarray} $ The coefficient on the $xy$ term is $1$ and the coefficient on the $y^2$ term is $-12$ , so to reverse the steps above, we need to find two numbers that add up to $1$ and multiply to $-12$ You can start by trying to guess which factors of $-12$ add up to $1$ . In other words, you need to find the values for $a$ and $b$ that meet the following conditions: $ {a} + {b} = {1}$ $ {a} \times {b} = {-12}$ If you're stuck, try listing out every single factor of $-12$ and its opposite as $a$ in these equations, and see if it gives a value for $b$ that validates both conditions. For example, since $3$ is a factor of $-12$ , try substituting $3$ for $a$ as well as $-3$ The two numbers $-3$ and $4$ satisfy both conditions: $ {-3} + {4} = {1} $ $ {-3} \times {4} = {-12} $ So we can factor the polynomial as $(x - 3y)(x + 4y)$.